Understanding PNP Transistor Working Modes: A Quick Question

[Bassalisk]

I want to ask a fairly simple question. Its about transistor working modes. Just to be sure I am thinking right.

If we apply positive voltages to both EB junction and CB junction. By that I mean that Positive potential comes to collector and positive potential comes to emitter. I've learned that this acts like a closed switch, right?

Does that mean that current will flow through transistor like through a piece of wire(ideally acting like a transistor wasn't even there)

And second case:

When transistor is directly polarized. And by that I mean EB in direct polarization and CB in inverse polarization.

Then we have current amplification. What does that mean actually?

Does that mean that I will have more current going through that wire than I would have if a transistor wasn't there?I hope I was clear with my question.

 

[vk6kro]

I think you have a wrong impression of how transistors function.

If you connect the collector of a PNP transistor to ground (the most negative voltage) and connect the emitter to the positive supply, no current will flow.

When you connect the base to ground (through a large resistor for safety), then a small current will flow from the base of the transistor through the resistor to ground.

The current that then flows from the positive supply and out of the collector will depend on the base current. It will be much larger than the base current but a fixed multiple of it.

This multiple is called the current gain.

None of this is any different to the operation of an NPN transistor, except that the polarity is reversed.

 

[Bassalisk]

[PLAIN]http://pokit.etf.ba/get/6248442d8e3a2e9e9039acb35fcbfd36.jpg

This is what I have in mind. How will this transistor behave?

 

[uart]

This is what I have in mind. How will this transistor behave?

Well as far as the external circuit is concerned there is NO transistor there, just a voltage source. So it's irrelevant.

 

[vk6kro]

There is a power source across the base-emitter junction with no limiting resistance.

So, if the voltage of this power source is more than 1 volt or so, the base-emitter junction will draw a very large current and be destroyed.

After that we don't have a transistor any more, but we do have two power supplies with probable short circuits on them.

 

[Bassalisk]

This picture helps with deriving formula for formulas for current components. Those voltages are just high enough to make both junctions directly polarized.

 

[uart]

But in your diagram the VI characteristic at the CE terminals is completely independent of the transistor. If you take the transistor out of the circuit nothing changes.

 

[Bassalisk]

But in your diagram the VI characteristics at the CE terminals is completely independent of the transistor. If you take the transistor out of the circuit nothing changes.

You mean this case will come to that? So this case will act like a closed switch?

Opposite to the case if those Sources were flipped and both junctions were in inverse bias and that would act like an open switch?

 

[vk6kro]

Ok, if there is enough voltage to just forward bias the base emitter junction and the two power supplies are equal, then there is no voltage between the collector and emitter.

The collector and emitter are equally positive relative to the base.

A transistor is not just two diodes. There is a path between the collector and the emitter that does not include the base. If this has no voltage across it then the transistor cannot operate as a transistor.

 

[Bassalisk]

I wish I could translate you my material, that I got this from.
[PLAIN]http://pokit.etf.ba/get/c2c35b60415e61b6ef0cc78fcf044010.jpg

This is mainly what I had in mind.

I was asking was this picture up top corresponding to saturation or closed switch.

 

[Dmytry]

Replace the voltage sources with current sources, that'll make more sense. The transistor will behave pretty much as two forward biased diodes indeed. The voltage on diodes will be the saturation voltage.

The situation you described can happen when you are driving an parallel LC oscillator, when you switch the transistor off. The LC oscillator starts oscillating and the collector-emitter voltage can reverse after one oscillation. In which case this can result in current flowing from base into collector.

 

[Bassalisk]

Replace the voltage sources with current sources, that'll make more sense. The transistor will behave pretty much as two forward biased diodes indeed. The voltage on diodes will be the saturation voltage.

The situation you described can happen when you are driving an parallel LC oscillator, when you switch the transistor off. The LC oscillator starts oscillating and the collector-emitter voltage can reverse after one oscillation. In which case this can result in current flowing from base into collector.

Can you explain or give me the link to LC oscillator with transistor? I want to study this further

 

[Dmytry]

Can you explain or give me the link to LC oscillator with transistor? I want to study this further

Well, it's a transient situation. Consider this:
http://en.wikipedia.org/wiki/Flyback_converter
The primary coil has stray capacitance between the leads. When you turn the switch off, the coil will oscillate, with voltage amplitude that can be larger than supply voltage, which would result in voltage reversal at the switch after full oscillation. If you use transistor as a switch there, you may end up in a situation when collector voltage falls below (npn) or rises above (pnp) the base voltage. In this case, base to collector junction becomes forward biased. I don't know any circuits where it is intended and used for anything productive though. I was just intending to illustrate how it can happen in the real world that base voltage of PNP transistor would be below collector and emitter voltage